\[h(\mathbf{x})=\mathbf{x}-\begin{pmatrix}\phantom{-} 1\\-21\end{pmatrix},\quad
g(\mathbf{x})=\begin{pmatrix}
98&-1\\1&\phantom{-} 98
\end{pmatrix}\mathbf{x},\quad
h^{-1}(\mathbf{x})=\mathbf{x}+\begin{pmatrix}\phantom{-}1\\-21\end{pmatrix}.\solnmarksplus{2}{1 for formulas for \(h\) and \(h^{-1}\),\\ 1 for formula for \(g\)\\ (can be implicit in working)}\]

Hence
\begin{align*}k(\mathbf{x})&=(h^{-1}\circ g \circ h)(\mathbf{x})\\&=h^{-1}(g(h(\mathbf{x})))\\
%line3
&=h^{-1}\left( g \left(\mathbf{x}-\begin{pmatrix}\phantom{-} 1\\-21\end{pmatrix}\right)\right)\\
%line4
&=h^{-1}\left(\begin{pmatrix}
98&-1\\1&\phantom{-} 98
\end{pmatrix}\left(\mathbf{x}-\begin{pmatrix} \phantom{-} 1\\-21\end{pmatrix}\right)\right)\solnmarksplus{1}{substituting formulas in correctly}\\
%line5
&=h^{-1}\left(\begin{pmatrix}
98&-1\\1&\phantom{-} 98
\end{pmatrix}\mathbf{x}-\begin{pmatrix}
98&-1\\1& \phantom{-} 98
\end{pmatrix}\begin{pmatrix} \phantom{-} 1\\-21\end{pmatrix}\right)\\
%line6
&=h^{-1}\left(\begin{pmatrix}
98&-1\\1& \phantom{-} 98
\end{pmatrix}\mathbf{x}-\begin{pmatrix}3\\1\end{pmatrix}\right)\solnmarksplus{1}{product of matrix and vector}\\
%line7
\end{align*}
